A random sample of 36 observations is drawn from a population with a mean equal to 66 and a standard deviation equal to 12. What is the mean and the standard deviation of the sampling distribution of x̄? μ = 66 σ = 2 Describe the shape of the sampling distribution of x̄. Does this answer depend on the sample size? The shape is that of a distribution and depend on the sample size. Calculate the z-score corresponding to a value of x̄ = 63.6. Calculate the z-score corresponding to a value of x̄ = 69.2. Find P(x̄ ≥ 63.6) (to 4 decimals) Find P(x̄ < 69.2) (to 4 decimals) Find P(63.6 ≤ x̄ ≤ 69.2) (to 4 decimals) There is a 60% chance that the value of x̄ is above (to 4 decimals).

μ = 66, σ = 2; The distribution is bell-shaped; Yes, this depends on the sample size; z = -1.2; z = 1.6; P(X ≥ 63.6) = 0.8849; P(X < 69.2) = 0.9452; P(63.6 ≤ X ≤ 69.2) = 0.8301; 65.5

Step-by-step explanation:

The central limit theorem states that if the sample size is greater than 30, the sample mean is roughly the same as the population mean. This means it is 66.

The standard deviation of a sampling distribution of means is given by

σ/√n

For our data, this is

12/(√36) = 12/6 = 2

The central limit theorem states that the sampling distribution is approximately normal, so it will be bell-shaped.

The formula for the z score of a sampling distribution of means is

[tex]z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}[/tex]

For the value of x = 63.6,

z = (63.6-66)/(12/(√36)) = -2.4/2 = -1.2

For the value of x = 69.2,

z = (69.2-66)/(12/(√36)) = 3.2/2 = 1.6

Using a z table, we see that the area under the curve to the left of z = -1.2 (for x = 63.6) is 0.1151. However, we want P(x̄ ≥ 63.6); this means we want the area to the right. We subtract our value from 1:

1-0.1151 = 0.8849

Using a z table, we see that the area under the curve to the left of z = 1.6 (for x = 69.2) is 0.9452. This is P(x̄ < 69.2).

Since we have the area under the curve to the left of each endpoint, to find P(63.6 ≤ x̄ ≤ 69.2) we subtract these values:

0.9452-0.1151 = 0.8301

To find the value that would correspond in 60% of values being larger than, we first consider the fact that the z table gives us areas to the left of values, which is probabilities less than the value. Our question is what number has a probability of 60% being larger than; this means we need to subtract from 1:

1-0.6 = 0.4

In a z table, we find the value as close to 0.4 as we can get. This is 0.4013, which corresponds with a z score of -0.25.

Substituting this into our z formula, we have

[tex]z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}\\\\-0.25=\frac{\bar{X}-66}{12\div \sqrt{36}}\\\\-0.25=\frac{\bar{X}-66}{12\div 6}\\\\-0.25=\frac{\bar{X}-66}{2}[/tex]

Multiply both sides by 2:

2(-0.25) = ((X-66)/2)(2)

-0.5 = X-66

Add 66 to each side:

-0.5+66 = X-66+66

65.5 = X